A pistoncylinder contains 3 lbm of air plus water vapor init

A piston-cylinder contains 3 lbm of air plus water vapor initially at 14.7 psi, 100 F, with a relative humidity of 70%. The mixture is cooled to 40 F.

a) Calculate the dew point of the mixture.

b) Calculate the amount of condensation (lbm)

Solution

We convert the unit system to SI and solve the problem and then convert back to FPS(English) Units.

a) Mass of mixture in piston-cylinder = 3 lbm = 1.3608 kg

Initial pressure =14.7 psi=101.4 kPa

Initial Temperature = 100 ^oF = 37.8 ^oC

Final temperature = 40 ^o F = 4.4 ^oC

RH = 70 %

From Psychrometric chart @ 37.8, 70 % RH the mixture contains 0.032 kg of water per kg of dry air

Let air quantity in the mixture be x

x(1+0.032) = 1.3608

Or air quantity x = 1.3186 kg, water vapour quantity = 0.0422 kg

Initial dew point temperature t_d1 = 33 ^o C

As the water will start condensing during cooling , the final dew point temperature is 4. 4 ^o C (same as final DBT) = 40 ^o F

b) From psychrometric chart, the final humidity ratio at 4.4 ^o C = 0.005 kg/kg of dry air.

Therefore, the water condensed m_w = m_a(W1 - W₂)

= 1.3186 X(0.032-0.005)

= 0.036 kg = 0.0794 lbm